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3 and 4 .Determinants and Matrices
easy
Let $A=\left(\begin{array}{cc}4 & -2 \\ \alpha & \beta\end{array}\right)$ . If $A ^{2}+\gamma A +18 I = O$, then $\operatorname{det}( A )$ is equal to
A
$-18$
B
$18$
C
$-50$
D
$50$
(JEE MAIN-2022)
Solution
The characteristic equation for $A$ is $| A -\lambda I |=0$
$\left|\begin{array}{cc}4-\lambda & -2 \\ \alpha & \beta-\lambda\end{array}\right|=0$
$(4-\lambda)(\beta-\lambda)+2 \alpha=0$
$\lambda^{2}-(\beta+4) \lambda+4 \beta+2 \alpha=0$
Put $\lambda= A$
$A^{2}-(\beta+4) A+(4 \beta+2 \alpha) I=0$
On comparison $-9(\beta+4)=\gamma \& 4 \beta+2 \alpha=18$
and $| A |=4 \beta+2 \alpha=18$
Standard 12
Mathematics
